3y^{''}-y^'+2y=0,y(0)=2,y^'(0)=0

Lösung

3 y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{\frac{t}{6}} (2 cos (\frac{23 t}{6}) - \frac{2}{23} sin (\frac{23 t}{6}))

Schritte zur Lösung

3 y^{''} (t) - y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{\frac{t}{6}} (2 cos (\frac{23 t}{6}) - \frac{2}{23} sin (\frac{23 t}{6}))

y = e^{\frac{t}{6}} (2 cos (\frac{23 t}{6}) - \frac{2}{23} sin (\frac{23 t}{6}))

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